Ratio — Part 1 MCQs with Explanations

$$\eqalign{ &\,\,\,\, {\text{ }}a{\text{ }}:{\text{ }}b{\text{ }}:{\text{ }}c \cr &\,\,\,\, {\text{ }}3{\text{ }}:{\text{ }}4{\text{ }}:{\text{ }}7{\text{ }} \cr & \boxed{\,\,3x:4x:7x\,\,\,\,} \Rightarrow 14x \cr & \therefore a + b + c = 14x \cr & c = 7x \cr & \therefore \left( {a + b + c} \right):c \cr & = 14x:7x \cr & = 2:1 \cr} $$

Let the number of students in the classes be 2x, 3x and 4x respectively; Total students = 2x + 3x + 4x = 9x $$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{the}}\,{\text{question}}, \cr & \frac{{ {2x + 12} }}{{ {3x + 12} }} = \frac{8}{{11}} \cr & or,\,24x + 96 = 22x + 132 \cr & or,\,2x = 132 - 96 \cr & or,\,x = \frac{{36}}{2} = 18 \cr & {\text{Hence,}} \cr & {\text{Original}}\,{\text{number}}\,{\text{of}}\,{\text{students}}, \cr & 9x = 9 \times 18 \cr & \,\,\,\,\,\,\,\, = 162 \cr} $$

Respective ratio of the NUMBER of coins; = 13 : 11 × 2 = 13 : 22 Hence, Number of 1 rupee coins; = $$\frac{{13 \times 210}}{{13 + 22}}$$ = 78

$$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{thequestion,}} \cr & {\frac{{2A}}{3}} = {\frac{{75B}}{{100}}} = {\frac{{C \times 6}}{{10}}} \cr & {\text{Above}}\,{\text{relation}}\,{\text{gives}}; \cr & \frac{{A \times 2}}{3} = \frac{{B \times 3}}{4} \cr & \to \frac{A}{B} = \frac{9}{8} \cr & {\text{And}}, \cr & \frac{{B \times 3}}{4} = \frac{{C \times 3}}{5} \cr & \to \frac{B}{C} = 4:5 \cr & \to \frac{B}{C} = 8:10 {\text{ (multiple by 2)}} \cr & {\text{Thus,}} \cr & A:B:C = 9:8:10 \cr} $$

$$\eqalign{ & {\frac{A}{B}} \times {\frac{B}{C}} = {\frac{3}{4}} \times {\frac{{12}}{{13}}} \cr & Or,\,\frac{A}{C} = \frac{{36}}{{52}} = 9:13 \cr} $$

Let A's Salary = Rs. 100 Then, B's Salary = Rs. 300 And, C's Salary = Rs. 400 Salary has given in 1 : 3 : 4 ratio Now, 5% increase in A's Salary, A's new Salary = (100 + 5% of 100) = Rs. 105 B's Salary increases by 10%, Then, B's new Salary = (300 + 10% of 300) = Rs. 330 C's Salary increases by 15%, C's new Salary = (400 + 15% of 400) = Rs. 460 Then, ratio of increased Salary, A : B : C = 105 : 330 : 460 = 21 : 66 : 92 Alternative 100 (A's salary) === 5%↑ ===> 105(A's increased salary) 300 (B's salary) === 10%↑ ===> 330 (B's increased salary) 400 (C's salary) === 15%↑ ===> 460 (C's increased salary) Ratio of their increased salary = 105 : 330 : 460 = 21 : 66 : 92

$$\eqalign{ & \frac{{\text{A}}}{{\text{B}}} = \frac{2}{3} \cr & \frac{{\text{B}}}{{\text{C}}} = \frac{4}{5} \cr} $$ A : B : C = 2 × 4 : 3 × 4 : 3 × 5 = 8 : 12 : 15

$$\eqalign{ & 2A = 3B \cr & Or,\,B = \left( {\frac{2}{3}} \right)A;\,{\text{and}} \cr & 2A = 4C \cr & Or,\,C = \left( {\frac{1}{2}} \right)A; \cr & {\text{Hence}}, \cr & A:B:C = A:\frac{{2A}}{3}:\frac{A}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1:\frac{2}{3}:\frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6:4:3 \cr} $$

Boys : girls = 8 : 5 (let the boys = 8x, girl = 5x) Total strength = 286 8x + 5x = 286 13x = 286 Or, x = $$\frac{{286}}{{13}}$$ = 22 Boys = 176 and girls = 110 22 more girls get admitted then number of girls become (5x + 22) = 110 + 22 = 132 Now, new ratio of boys and girls = 176 : 132 = 4 : 3

Let two numbers be 4x and 5x Their LCM = 180 and HCF = x Now, 1 st number × 2 nd number = LCM × HCF Or, 4x × 5x = 180 × x Or, 20x = 180 Or, x = 9 Then, the smaller number = 4 × 9 = 36

$$\eqalign{ & \frac{A}{D} = {\frac{A}{B}} \times {\frac{B}{C}} \times {\frac{C}{D}} \cr & \,\,\,\,\,\,\,\,\, = {\frac{2}{3}} \times {\frac{4}{5}} \times {\frac{5}{9}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{ {2 \times 4 \times 5} }}{{ {3 \times 5 \times 9} }} \cr & \,\,\,\,\,\,\,\,\, = \frac{8}{{27}} \cr & \,\,\,\,\,\,\,\,\, = 8:27 \cr} $$

Girls : boys = 6 : 5 Hence, girls = $$6 \times \frac{{66}}{{11}}$$ = 36 Boys = 30 New ratio, 30 : (36 + 4) = 3 : 4

$$\eqalign{ & {\text{Let }}x{\text{ be added to each term}}. \cr & {\text{According to question}}, \cr & \frac{{ {7 + x} }}{{11 + x}} = \frac{3}{4} \cr & Or,\,33 + 3x = 28 + 4x \cr & Or,\,x = 5 \cr} $$

Let the numbers be 7x and 11x. $$\eqalign{ & {\text{According to question}}, \cr & \frac{{ {7x + 7} }}{{ {11x + 7} }} = \frac{2}{3} \cr & Or,\,22x + 14 = 21x + 21 \cr & Or,\,x = 7 \cr & {\text{Smaller}}\,{\text{number}} = 49 \cr} $$

If we go through normal method, It will be quite cumbersome,so, We will solve this question through options Taking Option A: It has P + Q = 3. The possible value of $$\frac{{\text{P}}}{{\text{Q}}}$$ is $$\frac{1}{2}$$ or $$\frac{2}{1}$$ Using $$\frac{1}{2}$$, we see that on adding 2 in both the numerator and denominator we get $$\frac{3}{4}$$ (not required value) Similarly we check for $$\frac{2}{1}$$, this will also not give the required value Option B: We have $$\frac{1}{3}$$ possible ratio Then, we get the final value as $$\frac{3}{5}$$ (not = to $$\frac{1}{2}$$) Hence, rejected Option C: Here we have $$\frac{1}{4}$$ or $$\frac{2}{3}$$ Checking for $$\frac{1}{4}$$ we get $$\frac{3}{6}$$ = $$\frac{1}{2}$$ Hence, the option c is correct

Here,Let water = 7x and milk = 3x Now, 7x + 3x = 30 x = 3 So, water = 7x = 7 × 3 = 21 liter Milk = 3x = 3 × 3 = 9 liter Now, we keep milk constant and add water to mixture to get ratio 6 : 1 Let water in this mixture = 6y and milk = y We have, milk = 9 liter, so y = 9 liter Water = 6y = 6 × 9 = 54 liter Then extra water to be added is 33 liter

Let income of A and B be 3x and 2x respectively. Also, their expenditure is 5y and 3y. Now, according to question, 3x - 5y = 1000 ------- (i) × 3 2x - 3y = 1000 ---------- (ii) × 5 9x - 15y - 10x + 15y = 3000 - 5000 Or, -x = -2000 Or, x = 2000 Then, income of A = 3x = 3 × 2000 = Rs. 6000

Let the two positive numbers be 5x and 3x respectively According to question, 5x - 3x = 10 Or, x = 5Then numbers are 25 and 15 Thus, their product = 25 × 15 = 375

We will use work equivalence method, $$\eqalign{ & \frac{{30}}{{18}} = \frac{{ {\frac{1}{7}} }}{{ {\frac{6}{7}} }} \times \frac{x}{2} \cr & \frac{5}{3} = {\frac{1}{6}} \times \frac{x}{2} \cr & Or,\,x = \frac{{60}}{3} = 20\,{\text{days}} \cr} $$

Given; 3dog = 4 cat Or, $$\frac{{{\text{dog}}}}{{{\text{cat}}}} = \frac{4}{3}$$ Let cat's 1 leap = 3 meter and dogs 1 leap = 4 meter Then, ratio of speed of cat and dog = $$\frac{{3 \times 5}}{{4 \times 4}}$$ = 15 : 16

Let present age of A and B be 4x and 5x 18 years ago their ages $$\frac{{4x - 18}}{{5x - 18}} = \frac{{11}}{{16}}$$ Or, 64x - 288 = 55x - 198 Or, 64x - 55x = -198 + 288 Or, 9x = 90 Or, x = $$\frac{{90}}{9}$$ = 10 Sum of the present ages = 40 + 50 = 90 years

Quantity of milk and water in the 36 liter mixture 27 liter milk and 9 liter water Now, 15 liters milk is added then milk becomes 42 liters Now, ratio = 42 : 9 = 14 : 3

Let Maya's and Chhaya's present age is 6x and 5x respectively And, $$\frac{{6{\text{x}} + 15}}{{5{\text{x}} + 15}} = \frac{9}{8}$$ Or, 48x + 120 = 45x = 135 Or, 3x = 15 Or, x = 5 Present age of Maya = 6x = 30

Speed = $$24 - {\text{k}}\sqrt {\text{n}} $$ Putting the value, n = 4 we get, k = 2 Now the equation (as k = 2) become, S = $$24 - {\text{k}}\sqrt {\text{n}} $$ Thus, it means when n = 144, speed will be zero. Hence, train can just move when 143 wagons are attached

Given, x = y Or, x - y = 0 Or, (x - y) 2 = 0 Or, x 2 + y 2 - 2xy = 0 Or, x 2 + y 2 = 2xy It means that, x 2 + y 2 varies as xy

Let the number of passenger traveling by first class be x Then, number of passenger traveling by second class will be 4x But the fare is in the ratio 3 : 1 In other words, if 3y fare is collected per I class passenger, y would be collected per II class passenger Fares of I class passengers : Fares of II class passengers = x × 3y : 4x × y = 3 : 4 The above ratio can be interpreted as follows If total fare is 3 + 4 = 7, then I class passengers should pay Rs. 3 Similarly, we can calculate the fare of I class passengers when total was 224000 Total Fare Class Fare 7 3 224000 ? = $$224000 \times \frac{3}{7}$$ = Rs. 96000

Amount of alcohol in first vessel, = 25% of 2 litre = 0.25 × 2 = 0.5 litre Amount of alcohol in second vessel, = 40% of 6 litre = 0.4 × 6 = 2.4 litre Total amount of alcohol out of 10 litres of mixture is 0.5 + 2.4 = 2.9 litre Thus, Concentration of the mixture is, $$\frac{{2.9 \times 100}}{{10}}$$ = 29%

Let, B 1 : B 2 : B 3 = 3x : 4x : 5x and B 1 : B 2 : B 3 = 5y : 4y : 3y Number of oranges remain constant in third basket as increase in oranges takes place only in first two baskets. Hence, 5x = 3y x = $$\frac{3y}{5}$$ and, ∴ 3x : 4x : 5x (putting the vale of x) = $$\frac{{9{\text{y}}}}{5}:\frac{{{\text{12y}}}}{5}:\frac{{{\text{15y}}}}{5}$$ = 9y : 12y : 15y And, 5y : 4y : 3y (multiple by 5) → 25y : 20y : 15y ∴Increment in first basket = 16 And, Increment in second basket = 8 Thus, Required ratio = $$\frac{{16}}{8}$$ = 2 : 1

Maximum earning will be only possible when he will won on the maximum yielding table A → 10 : 1 B → 20 : 1 C → 30 : 1 i.e., he won B and C but lost on A 20 × 200 + 30 × 200 - 1 × 200 = 9800 Minimum earnings will be when he won on table A and B and lose on table 3 10 × 200 + 20 × 200 - 1 × 200 = 5800 Therefore, difference = 9800 - 5800 = Rs. 4000 Alternatively, The difference, = [(30 + 20 - 1) - (10 + 20 -1)] × 200 = Rs. 4000

Speed of track = 550 per minute. Speed of bus = $$\frac{{33\,{\text{kms}}}}{{45}}$$ = $$\frac{{33000}}{{45}}$$ = 733.33 m/minute Ratio of their speeds = $$\frac{{550}}{{733.33}}$$ = 3 : 4

Ratio of number of men, women and children, $$\eqalign{ & = \frac{{18}}{6}:\frac{{10}}{5}:\frac{{12}}{3} \cr & = 3:2:4 \cr} $$ Total (Men + Women + Children) = 18 3X + 2X + 4X = 18 9X = 18 X = 2 Hence, number of women = 2X = 2 × 2 = 4 Share of all women = $$\frac{{10 \times 4000}}{{40}}$$ = Rs. 1000 [18 + 10 + 12 = 40] Thus, share of each woman = $$\frac{{1000}}{4}$$ = Rs. 250

Ratio of Gold and Copper in Alloy A = 5 : 3 Ratio of Gold and Copper in Alloy B = 5 : 11 Amount of Gold in Alloy A = $$\frac{5}{8}$$ Amount of Gold In Alloy B = $$\frac{5}{{16}}$$ Amount of Copper in A = $$\frac{3}{8}$$ Amount of Copper in B = $$\frac{{11}}{{16}}$$ Amount of Gold In C, = (Amount of gold in A + Amount of gold in B) = $$\frac{5}{8}$$ + $$\frac{5}{{16}}$$ = $$\frac{{10 + 5}}{{16}}$$ = $$\frac{{15}}{{16}}$$ Amount of Copper in C, = Amount of Copper in A + Amount of Copper in B = $$\frac{3}{8}$$ + $$\frac{{11}}{{16}}$$ = $$\frac{{17}}{{16}}$$ So, Ratio of Gold and Copper in C, $$ = \frac{{15}}{{16}}:\frac{{17}}{{16}} = 15:17$$

Let X coins of each type of there Total Value = Rs. 35 Now, X + $$\frac{{\text{X}}}{2}$$ + $$\frac{{\text{X}}}{4}$$ = 35 4X + 2X + X = 140 7X = 140 X = 20

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively. When 9 litres of mixture are drawn off, quantity of A in mixture left: $$\eqalign{ & = \left[ {7x - {\frac{7}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {7x - {\frac{{21}}{4}} } \right]\,{\text{litres}} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & = \left[ {5x - {\frac{5}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {5x - {\frac{{15}}{4}} } \right]\,{\text{litres}} \cr & \therefore \,{\text{ratio becomes}}, \cr & \frac{{ {7x - {\frac{{21}}{4}} } }}{{ {\left(5x - {\frac{{15}}{4}}\right)+9 } }} = \frac{7}{9} \cr & \Rightarrow \frac{{ {28x - 21} }}{{ {20x + 21} }} = \frac{7}{9} \cr & \Rightarrow {252x - 189} = 140x + 147 \cr & \Rightarrow 112x = 336 \cr & \Rightarrow x = 3 \cr & {\text{So the can contained}}, \cr & = 7 \times x \cr & = 7 \times 3 \cr & = 21\,{\text{litres of A initially}}{\text{.}} \cr} $$

Let bucket contains 5x and 3x of liquids A and B respectively. When 16 litres of mixture are drawn off, quantity of A in mixture left: $$\eqalign{ & {5x - {\frac{5}{8}} \times 16} = {5x - 10} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & {3x - {\frac{3}{8}} \times 16} = {3x - 6} \cr & {\text{Now the ratio becomes,}} \cr & \frac{{ {5x - 10} }}{{ {3x - 6} }} = \frac{3}{5} \cr & \Rightarrow 25x - 50 = 9x - 18 \cr & \Rightarrow 16x = 32 \cr & \Rightarrow x = 2 \cr & {\text{So, quantity of liquid B initially}}, \cr & = 3 \times 2 = 6\,{\text{litres}} \cr} $$

Let the quantity of the wine in the cask originally be x litres. Using formula: Final Amount of solute that is not replaced = Initial Amount × $${\left( {\frac{{{\text{Vol}}{\text{. after removal}}}}{{{\text{Vol}}{\text{. after replacing}}}}} \right)^{\text{N}}}$$ Where N = No. of operation done. Then ratio of wine to total solution in cask after 4 operations, $$\eqalign{ & 1 \times { {\left( {\frac{{x - 8}}{x}} \right)} ^4} = \frac{{16}}{{81}} \cr & \Rightarrow 1 \times {\left\{ {\frac{{ {x - 8} }}{x}} \right\}^4} = {\left( {\frac{2}{3}} \right)^4} \cr & \Rightarrow \frac{{ {x - 8} }}{x} = \frac{2}{3} \cr & \Rightarrow 3x - 24 = 2x \cr & \Rightarrow x = 24\,{\text{litres}} \cr} $$

Milk : Water 7 : 5 7 : 8 3 unit ∴ Remember water is added and not milk, so make milk equal but here milk is already equal 3 units = 15 litres 1 units = 5 litres 8 units = 40 litres Total quantity of water in the new mixture = 40 litres

$$\frac{{\text{x}}}{{\text{y}}} = \frac{5}{2}$$ Means x = 5, y = 2 Putting value of x and y in expression 8 × 5 + 9 × 2 = 58 8 × 5 + 2 × 2 = 44 58 : 44 = 29 : 22

Given; 7 jumps of Tom = 5 jumps of Jerry Or, $$\frac{{{\text{Tom}}}}{{{\text{Jerry}}}} = \frac{5}{7}$$ Let Jerry's 1 leap = 7 meter and Tom's 1 leap = 5 meter Then, ratio of speed of Tom and Jerry = $$\frac{{8 \times 5}}{{6 \times 7}}$$ = $$\frac{{40}}{{42}}$$ = 20 : 21

Ratio of Ducks and Frogs in Pond = 37 : 39 Average of Ducks and Frogs in Pond = 152 So, total number of Ducks and Frogs in the Pond = 2 × 152 = 304 ∴ Number of Frogs = $$\frac{{304 \times 39}}{{76}}$$ = 156